Python dictionary operations

1. Define a dictionary

>>> D1 = {//key-value format

"Student1": "Xiaoming",

"Student2": "Zhangsan",

"Student3": "Lisi"

}

2. View

>>> print (d1["student1"])//used with the known key, error when this key is not available

Xiaoming

>>> Print (D1.get ("Student2"))//get method does not error

Zhangsan

>>> Print (D1.get ("Student1"))

None

3. Change the value of key

>>> print (d1["Student1"])

Xiaoming

>>> d1["student1"] = "xiaoming"//Modify if present, add if not present

>>> print (d1["Student1"])

Xiaoming

4. Delete

>>> del d1["Student1"]

>>> Print (D1)

{' Student2 ': ' Zhangsan ', ' student3 ': ' Lisi '}

5. Judge if there is any data in the dictionary

>>> print ("Student1" in D1)

False

6, Multi-level dictionary nesting

>>> D2 = {

"Weibo": {

"Guomao": ["Wilson", "David", "Lucia"],

"Bagou": ["Zhangsan", "Wangwu", "Lisi"]},

"Xindongfang": {

"Yuquan": ["Piter", "Alen", "CC"],

"Wukesong": ["Shenyang", "Lidong", "Wangwei"]}

}

7. Modify the values in the nested menu

>>> print (d2["Weibo" ["Guomao"])

[' Wilson ', ' David ', ' Lucia ']//before modification

>>> d2["Weibo" ["Guomao"][1] = "David"

>>> print (d2["Weibo" ["Guomao"])

[' Wilson ', ' David ', ' Lucia ']//after modification

8. Print all values in the dictionary

>>> print (D2.values ())

9. Print all keys in the dictionary

>>> print (D2.keys ())

10, if the dictionary exists this key is returned, does not exist to create

>>> Print (D2)

{' Weibo ': {' Guomao ': [' Wilson ', ' David ', ' Lucia '], ' Bagou ': [' Zhangsan ', ' Wangwu ', ' Lisi ']}, ' Xindongfang ': {' Yuquan ': [' Pit Er ', ' Alen ', ' cc '], ' wukesong ': [' Shenyang ', ' Lidong ', ' Wangwei '}}

>>> D2.setdefault ("Weibo", {"Xinhua": ["AA", "BB", "CC"]})

{' Guomao ': [' Wilson ', ' David ', ' Lucia '], ' Bagou ': [' Zhangsan ', ' Wangwu ', ' Lisi ']}

>>> D2.setdefault ("Yingfu", {"Xinhua": ["AA", "BB", "CC"]})

{' Xinhua ': [' AA ', ' BB ', ' CC ']}

>>> Print (D2)

{' Weibo ': {' Guomao ': [' Wilson ', ' David ', ' Lucia '], ' Bagou ': [' Zhangsan ', ' Wangwu ', ' Lisi ']}, ' Xindongfang ': {' Yuquan ': [' Pit Er ', ' Alen ', ' cc '], ' wukesong ': [' Shenyang ', ' Lidong ', ' Wangwei ']}, ' Yingfu ': {' Xinhua ': [' AA ', ' BB ', ' CC '}}

11, merge two dictionaries, if the dictionary has the same key is updated, no add

>>> D1

{' Student2 ': ' Zhangsan ', ' student3 ': ' Lisi '}

>>> D3 = {"Student2": "AAAA", "bbbbb": "CCCCC", 1:2}

>>> D1.update (D3)

>>> Print (D1)

{' Student2 ': ' AAAA ', ' student3 ': ' Lisi ', ' bbbbb ': ' CCCCC ', 1:2}

12. Convert Dictionaries to lists

>>> print (D1.items ())

Dict_items ([' Student2 ', ' AAAA '), (' Student3 ', ' Lisi '), (' bbbbb ', ' CCCCC '), (1, 2)])

Can cycle print

>>> for K,v in D1.items ():

.. print (K,V)

...

Student2 AAAA

Student3 Lisi

BBBBB CCCCC

1 2

13, the Dictionary of the Cycle

>>> for I in D1:

... print (i)//only key

...

Student2

Student3

bbbbb

1

>>> for I in D1:

... print (i,d1[i])//Pass I as key into the dictionary query

...

Student2 AAAA

Student3 Lisi

BBBBB CCCCC

0 S

14, for K,v in D1.items (): And for I in D1:print (I,d1[i]) difference

The second way is more efficient, because the first way to convert the dictionary to a list, in the case of large amounts of data, will lead to waste of resources.

Python dictionary operations